Finding Area of a Right Triangle & Proving arctanx Limit

[azatkgz]

Question1:In a right triangle,the sum of the lengths of the hypotenuse and a leg is 1.Find the largest possible area of such triangle.
Question2:Prove that arctanx=pi/2-1/x+o(1/x)(x tends to +inf)
Question3:evaluate lim(2arctanx/pi)^2 (x tends to +inf)

 

[sutupidmath]

well for the first one here it is my approach, i hope i am not wrong.
let c be the hypotenuse, and a, b to legs of that right triangle. Let us take
c+b=1 ----------*
we know that the area of a right triangle is

A=(ab)/2, from * we have A=(a/2)(1-c),, so this is a function, and we need to find its maximum. finding the first derivative we get A'=1-c-a, so the maximum of this function is when a=1-c, so the maximum area is A=(b^2)/2, or (1-c)^2/2

the tird one, i guess the limit is 2pi, i did not do any calculations though, just glanced it.
and for the second one, try to expand arctanx using the taylor formula, when n=2, i mean just applying the derivative twice.
i hope i was helpful.

 

[tacman]

Question1: To find the largest possible area of a right triangle, we can use the formula A = 1/2 * base * height. In this case, the base and height are the two legs of the triangle. Let's call the hypotenuse x and the leg y. We know that x + y = 1, so we can rewrite the area formula as A = 1/2 * y * (1 - y). To find the maximum value of A, we can take the derivative and set it equal to 0:

A' = 1/2 - y = 0
y = 1/2

Substituting y = 1/2 back into the area formula, we get A = 1/2 * 1/2 * (1 - 1/2) = 1/8. Therefore, the largest possible area of the right triangle is 1/8 when the hypotenuse is 1/2 and one of the legs is also 1/2.

Question2: To prove that arctanx = pi/2 - 1/x + o(1/x) as x tends to +inf, we can use the definition of the arctan function:

arctanx = tan^-1(x) = y if and only if tan(y) = x

As x tends to +inf, the value of y also tends to +inf. Therefore, we can use the following limit property:

lim (tan^-1(x)) = pi/2 - 1/x + o(1/x) as x tends to +inf

Question3: To evaluate lim (2arctanx/pi)^2 as x tends to +inf, we can use the limit properties of trigonometric functions:

lim (2arctanx/pi)^2 = (2pi/2)^2 = pi^2 as x tends to +inf

Therefore, the limit of (2arctanx/pi)^2 as x tends to +inf is pi^2.